tough question

[sana.noor]

At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E

[mkdureja]

From 1PM to 1.30 PM, only Train X covers the distance between the two trains.
Its speed = r miles/hr. So, distance it covers in 1/2 hr. is r/2 miles.

Now, from then on, both the trains cover the remaining distance between them 'p-r/2' miles(subtracting the distance covered by Train X in the first 1/2 hr. from the total distance), with the speed of 'r+s' miles/hr (combined speed).

So, time it will take them to reduce their distance to zero will be =
(p-r/2)/(r+s).

Add initial 1/2 hr. to get the answer = 1/2 + (p-r/2)/(r+s).

Hope its clear.

[GMATGuruNY]

At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E

Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.

[subhakam]

At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E

Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.

Hello GMAT GuruNY - can you please comment whether what my algebraic approach is correct ? (Need to know where I am going wrong)

Let us say we need to fine time (t) when the two trains meet after 1 pm

For Train X - speed is r miles per hour
since it leaves at 1 pm it has 30 minutes head start over train y , therefore total time it gets to travel is (t+1/2)
For Train Y - speed is s miles per hour and time it travels is t
Now , both of them cover the given distance p , therefore equation is
r(t+1/2) + st = p => [r{(2t+1)}/]2 + st = p
=> 2rt+ r + 2st = 2p
=> 2rt + 2st = 2p-r => 2t(r+s)= 2p-r
=> t = [p-r/2] / (r+s) = option D
However do we need to add 1/2 more since travel by x is actually at 1 pm and total time traveled by train X = actually t +1/2 ? in other words
1/2 + ([p-r/2]/(r+s)) ? Let me know
In the same way could one of the right options ever be t = ([p+s/2]/(r+s))- 1/2 ?? (if we say from point of X time traveled is t and from point of Y time traveled is (t-1/2)i.e. train Y travels 30 minutes less than that traveled by train X ?
Please let me know - i want to get this right conceptually and algebraically
Thanks
Subhakam

[Atekihcan]

Let us say we need to fine time (t) when the two trains meet after 1 pm

For Train X - speed is r miles per hour
since it leaves at 1 pm it has 30 minutes head start over train y , therefore total time it gets to travel is (t+1/2)
For Train Y - speed is s miles per hour and time it travels is t

As you are assuming that t is the time in hours after 1 PM when they meet, train X has traveled for t hours and train Y has traveled for (t - 1/2) hours.

You are assuming that t is the time in hours after 1 PM when they meet, but you are doing the calculation assuming that t is the time in hours after 1:30 PM when they meet.

So, in our final answer you need to add 1/2.

[[email protected]]

At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]


Why have we taken 2 as the rate of Y can you please help me understand!


OA is E

Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.

[GMATGuruNY]

At 1:00 PM, Train X departed from Station A on the road to Station B. At 1:30 PM, Train Y departed Station B on the same road for Station A. If Station A and Station B are p miles apart, Train XÂ’s speed is r miles per hour, and Train Y'Â’s speed is s miles per hour, how many hours after 1:00 PM, in
terms of p, r, and s, do the two trains pass each other?
(A) [1/2 + p-(s/2)]/r+s
(B) (p-s/2)/r+s
(C) 1/2 + [(p-r/2)/r]
(D) [p-(r/2)]/r+s
(E) 1/2 + [p-(r/2)/r+s]

OA is E

Let p = 10 miles (the total distance), r = 20 miles per hour (X's rate), and s= 2 miles per hour (Y's rate).
In the 30 minutes from 1-1:30pm, the distance traveled by X = r*t = 20*(1/2) = 10 miles.
Since X travels the ENTIRE DISTANCE and meets Y by 1:30pm, the total amount of time required for X and Y to meet = 1/2.
This is our target.

Now we plug p=10, r=20, and s=2 into the answers to see which yields our target of 1/2.
A quick scan reveals that only C and E are viable:
C: 1/2 + [(p-r/2)/r] = 1/2 + 0 = 1/2.
E: 1/2 + [p-(r/2)/r+s] = 1/2 + 0 = 1/2.

The correct answer choice must include the value of s, since Y's rate will affect the total time if X does NOT travel the entire distance by 1:30pm.
Eliminate C.

The correct answer is E.

Why have we taken 2 as the rate of Y can you please help me understand!

When p=10 and r=20, train Y does not travel any portion of the distance.
Thus, we can plug in ANY VALUE for s (train Y's speed).
I chose s=2 because small values typically are easier to evaluate.
If we plug in p=10, r=20, and s=1000, the result is the same: only C and E yield the target value (1/2).